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M06-16

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M06-16  [#permalink]

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New post 16 Sep 2014, 00:27
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A
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D
E

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If \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\) and \(t\), \(u\), \(x\), and \(y\) are non-zero integers, which of the following is true?

A. \(\frac{t}{u}=1\)
B. \(\frac{y}{x}=-1\)
C. \(t = -u\)
D. \(t = \pm u\)
E. None of the above

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Re M06-16  [#permalink]

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New post 16 Sep 2014, 00:27
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Official Solution:

If \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\) and \(t\), \(u\), \(x\), and \(y\) are non-zero integers, which of the following is true?

A. \(\frac{t}{u}=1\)
B. \(\frac{y}{x}=-1\)
C. \(t = -u\)
D. \(t = \pm u\)
E. None of the above


Given that: \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{y} = \frac{u}{x}\).

So, \(\frac{t}{u} = \frac{x}{y}\) and \(\frac{t}{u} = \frac{y}{x}\) (from 2), which means that \(\frac{t}{u}\) and \(\frac{x}{y}\) equal to their reciprocals: \(\frac{t}{u}=\frac{u}{t}\) and \(\frac{x}{y}=\frac{y}{x}\). So, \(t^2=u^2\) and \(t^2=u^2\), which gives \(|t|=|u|\) (or which is the same \(t = \pm u\)) and \(|x|=|y|\) (or which is the same \(x = \pm y\)).


Answer: D
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Re: M06-16  [#permalink]

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New post 13 Feb 2015, 22:18
1
Eqn 1....Let t/u = x/y = k

Therefore t = uk, x = yk.....

Substitute in eqn 2...t/y = u/x

uk/y = u/yk

k^2 = 1

k = +/-1

so t = +/- u
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Re: M06-16  [#permalink]

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New post 17 Feb 2015, 11:55
Bunuel In questions where they ask which is true.. are they asking for must be true or which can be true?

Asking since i solved this question as which can be true and stopped after choosing A as A was coming as true for a case where all numbers were equal.
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Re: M06-16  [#permalink]

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New post 17 Feb 2015, 11:57
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qw1981 wrote:
Bunuel In questions where they ask which is true.. are they asking for must be true or which can be true?

Asking since i solved this question as which can be true and stopped after choosing A as A was coming as true for a case where all numbers were equal.


Which is true = which must be true.
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Re: M06-16  [#permalink]

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New post 10 Jun 2015, 11:54
Why are we cross-multiplying ? The question does not state that integers t,x,u,v are positive...
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Re: M06-16  [#permalink]

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New post 10 Jun 2015, 11:56
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Re: M06-16  [#permalink]

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New post 11 Dec 2015, 05:31
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?
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Re: M06-16  [#permalink]

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New post 26 Apr 2016, 18:39
m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?


t=u
would be true for negative and positive values of u.
But I can see that some people would choose E.
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Re: M06-16  [#permalink]

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New post 15 Jun 2016, 02:34
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m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?



One explanation that i can offer here is as follows -

The equation t(x+y)=u(x+y) holds good in almost all case, however (given x,y are non zero integers), consider a scenario where x= -y.
In this case you cannot say that t=u as you cant divide both sides with (x+y) !!! why? coz its 0 :)

Incidentally this is one of the wrong answer choices as well
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Re: M06-16  [#permalink]

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New post 18 Feb 2017, 07:53
I still can't understand why option D is correct, even though the question is MUST BE TRUE.
Here is my thinking:

Let's say t, u, x and y are all equal to 1 (non-zero integer). Then according to option D: 1 = +-1, which is only true for positive sign, but not negative. I mean the option D would be a MUST BE TRUE answer only if it was written like this:
t = u OR t = -u.
As far as I understand t = +-u means 't = u AND t = -u'
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Re: M06-16  [#permalink]

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New post 10 Jul 2018, 19:42
why is |t| = |u| the same thing as t = ±u? I don't understand that. Can someone please explain?
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New post 10 Jul 2018, 20:27
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Re: M06-16  [#permalink]

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New post 13 Aug 2018, 07:24
m9338 wrote:
Hi,

I did the following:

From (1), ty=xu (cross multiplying)
From (2), tx=uy

so adding each side of the equalities, t(x+y)=u(x+y) so t=u (and not t= +-u)

What am I doing wrong?


I still cannot understand why this is wrong. Could someone help me?
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Re: M06-16  [#permalink]

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New post 05 Sep 2018, 06:40
So, t^2=u^2 and t^2=u^2 gives |t|=|u|

Could someone please elaborate the result derived above?
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Re: M06-16  [#permalink]

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New post 05 Sep 2018, 06:45
Megha1119 wrote:
So, t^2=u^2 and t^2=u^2 gives |t|=|u|

Could someone please elaborate the result derived above?


By taking the square root.

Remember: \(\sqrt{x^2}=|x|\).

The point here is that square root function can not give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=|x|\)
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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